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%I #31 Mar 20 2018 01:44:23
%S 1,0,0,0,1,0,1,0,0,1,0,0,1,0,0,0,1,1,1,1,0,1,0,1,2,1,0,1,0,0,0,1,1,0,
%T 0,0,0,1,0,0,2,0,0,1,0,1,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,
%U 0,0,1,1,1,1,0,0,1,0,1,0,0,1,0,1,2,1,0,1,0,0,1,0,0,1,0,0,0,1,1
%N S(m,n) = (number of nonzero terms common to the base 3 expansions of m and n), a symmetric matrix read by antidiagonals.
%C Every nonnegative integer occurs infinitely many times in the matrix.
%H Robert Israel, <a href="/A206567/b206567.txt">Table of n, a(n) for n = 1..10011</a> (antidiagonals 1 to 141, flattened)
%F Diagonal entries S(n,n) = A160384(n) since all nonzero digits match. - _Robert Israel_, Mar 18 2018
%e Northwest corner:
%e 1 0 0 1 0 0 1 0 0 1 0 0 1
%e 0 1 0 0 1 0 0 1 0 0 1 0 0
%e 0 0 1 1 1 0 0 0 0 0 0 1 1
%e 1 0 1 2 1 0 1 0 0 1 0 1 2
%e 0 1 1 1 2 0 0 1 0 0 1 1 1
%e 0 0 0 0 0 1 1 1 0 0 0 0 0
%e 1 0 0 1 0 1 2 1 0 1 0 0 1
%e 0 1 0 0 1 1 1 2 0 0 1 0 0
%e 0 0 0 0 0 0 0 0 1 1 1 1 1
%e 1 0 0 1 0 0 1 0 1 2 1 1 2
%e 0 1 0 0 1 0 0 1 1 1 2 1 1
%e 0 0 1 1 1 0 0 0 1 1 1 2 2
%e 1 0 1 2 1 0 1 0 1 2 1 2 3
%e 4 = 3 + 1 and 13 = 3^2 + 3 + 1, so S(13,4)=2.
%p S:= proc(m,n) local M,N;
%p M:= convert(m,base,3);
%p N:= convert(n,base,3);
%p convert(zip((s,t) -> `if`(s=t and s <> 0, 1, 0),M,N),`+`);
%p end proc:
%p seq(seq(S(k,n-k+1),k=1..n),n=1..30); # _Robert Israel_, Mar 19 2018
%t d[n_] := IntegerDigits[n, 3];
%t t[n_] := Reverse[Array[d, 100][[n]]]
%t s[n_, k_] := Position[t[n], k]
%t t[m_, n_] := Sum[Length[Intersection[s[m, k], s[n, k]]], {k, 1, 2}]
%t TableForm[Table[t[m, n], {m, 1, 24},
%t {n, 1, 24}]] (* A206567 as a matrix *)
%t Flatten[Table[t[i, n + 1 - i], {n, 1, 24},
%t {i, 1, n}]] (* A206567 as a sequence *)
%o (PARI) d(n) = Vecrev(digits(n, 3));
%o T(n, k) = {my(dn = d(n), dk = d(k), nb = min(#dn, #dk)); sum(i=1, nb, dn[i] && (dn[i] == dk[i]));} \\ _Michel Marcus_, Mar 19 2018
%Y Cf. A160384, A206479 (similar in base 2).
%K nonn,tabl,base
%O 1,25
%A _Clark Kimberling_, Feb 09 2012
%E Edited by _Robert Israel_, Mar 19 2018
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